Rectangle ABCD
A is (-1,4), B is (4,8)
Equation of BC is 4y + 5x= 52
equation of AD is 4y + 5x = 11
equation of AB is 5y= 4x+ 24
Find C and D
Answer is (12,-2) and (7,-6)
I've tried gradient,length,midpoint formulas but to no avail.
Thanks
Rectangle ABCD
A is (-1,4), B is (4,8)
Equation of BC is 4y + 5x= 52
equation of AD is 4y + 5x = 11
equation of AB is 5y= 4x+ 24
Find C and D
Answer is (12,-2) and (7,-6)
I've tried gradient,length,midpoint formulas but to no avail.
Thanks
And, you now tell us, "Length of BC is twice the length Of AB"
It is easy to calculate that the length of AB is $\displaystyle \sqrt{(4-(-1)^2+ (8- 4)^2}= \sqrt{25+ 16}= \sqrt{41}$
so the length of BC is $\displaystyle 2\sqrt{41}$ and, because this is a rectangle, BC and AD are the same length.
Take D= (x, y)= (x, (11- 5x)/4). Then we must have [tex]\sqrt{(x- (-1))^2+ ((11- 5x)/4- 4)^2}= 2\sqrt{41}[tex]
so that, squaring both sides gives $\displaystyle (x+1)^2+ (5x+ 5)^2/16= 164$. Solve that quadratic equation for x. Do the same with BC.
Find C and D
Answer is (12,-2) and (7,-6)
I've tried gradient,length,midpoint formulas but to no avail.
Thanks
given the two points and the fact that the length of the rectangle is twice width(AB) there is a simple method to find the other vertices
slope diagram AB = 4/5
slope diagram AD =-5/4
actual slope diagram AD =-10/8 since AD =2AC
Point D x =8-1 =7 y =4 + absolute value y =10 y=-6
same procedure gives point C